NCERT Solutions for Class 11 Chemistry Chapter 6: Thermodynamics

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A thermodynamic state function is a quantity:

(a) used to determine heat changes

(b) whose value is independent of path

(c) used to determine pressure volume work

(d) whose value depends on temperature only.

Ans. (b) whose value is independent of a path.

Explanation:

The property which depends on the state of a system and independent of a path followed to attain it, is called as a state function. Thus, a thermodynamic state function is a quantity whose value is independent of a path’.

6.2. For the process to occur under adiabatic conditions, the correct condition is:

(a) ΔT = 0

(b) Δp = 0

(c) q = 0

(d) w = 0

Ans. (c) q = 0

Explanation:

Adiabatic process is a process in which there is no transfer of heat between the system and surroundings. Thus, a system is said to be under adiabatic conditions if there is no exchange of heat (q) between the system and its surroundings, i.e., q = 0.

6.3. The enthalpies of all elements in their standard states are:

(a) unity

(b) zero

(c) < 0

(d) different for each element

Ans. (b) zero

Explanation:

By definition, the enthalpy of formation of elements in their standard state is taken as zero.

Therefore, the enthalpies of all elements in their standard states is zero irrespective of the element.

6.4. ΔU0 of combustion of methane is – X kJ mol–1. The value of ΔH0 is:

(a) = ΔU0

(b) > ΔU0

(c) < ΔU0

(d) = 0

Ans. (c) < ΔU0
Explanation:

$$\text{CH}_4\text{(g)}+2\text{O}_2\text{(g)}\xrightarrow{}\text{CO}_2\text{(g)}+2\text{H}_2\text{O}\text{(g)}$$

Δn(g) = nP – nR

= 1 – 3 = – 2

ΔH° = ΔU° + Δn(g) RT

ΔH° = ΔU° – 2RT

ΔH° + 2RT = ΔU°

ΔU° > ΔH°

6.5. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy of formation of CH4(g) will be :

(a) –74.8 kJ mol–1

(b) –52.27 kJ mol–1

(c) +74.8 kJ mol–1

(d) +52.26 kJ mol–1

Ans. (a) –74.8 kJ mol–1

Explanation:

According to the question,

$$\text{CH}_4\text{(g)}+2\text{O}_2\text{(g)}\xrightarrow{}\text{CO}_2\text{(g)}+2\text{H}_2\text{O}\text{(l)}\\\Delta \text{H}=-890.3\space\text{kJ mol}^{\normalsize–1}\space\text{...(i)}\\\text{C(s)+O}_2(g)\xrightarrow{}\text{CO}_2(g)\\\Delta \text{H}=-393.5\space\text{kJ}\space\text{mol}^{\normalsize -1}\space\text{...(ii)}\\\text{2H}_2\text{(g)}+\text{O}_2\text{(g)}\xrightarrow{}\text{2H}_2\text{O}\space\text{(g)}\\\Delta\text{H}=–285.8 \text{kJ mol}^{\normalsize –1} \text{...(iii)}$$

On multiplying equation (iii) by 2 and adding with equation (ii) and subtracting equation (i) from it, we get the desired equation,

$$\text{C(s) + 2H}_2\text{(g)}\xrightarrow{}\text{CH}_4\text{(g)}$$

Therefore, enthalpy of formation of methane is:

$$\text{C(s) + 2H}_2(g)\xrightarrow{}\text{CH}_4\text{(g)}$$

fHCH4 = ΔcHc+2ΔcHcHH2cHco2

= [–393.5 + 2(–285.8) – (–890.3)] kJ mol–1

= – 74.8 kJ mol–1

Thus, enthalpy of formation of methane is –74.8 kJ mol–1.

6.6. A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be

(a) possible at high temperature

(b) possible only at low temperature

(c) not possible at any temperature

(d) possible at any temperature

Ans. (d) possible at any temperature

Explanation:

For a reaction to be spontaneous, DG should be negative.

ΔG = ΔH – TDS

For the given reaction,

ΔS = positive

And ΔH = negative (since heat is evolved)

and ΔG = negative

Therefore, the reaction is spontaneous at any temperature.

6.7. In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

Ans. According to the first law of thermodynamics,

ΔU = q + W ...(i)

Where,

ΔU = Change in internal energy

q = Heat = 701 J (Heat is absorbed so, it is positive) (Given)

W = Work done = ( –394 J) (Work is done by the system so, it is negative) (Given)

Substituting the values in expression (i), we get

ΔU = 701 + (–394)

ΔU = 307 J

Thus, the change in internal energy for the given process is 307 J.

6.8. The reaction of cyanamide, NH2CN(s), with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be –742.7 kJ mol–1 at 298 K. Calculate enthalpy change for the reaction at 298 K.

$$\textbf{NH}_2\textbf{CN(g)}+\frac{\textbf{3}}{\textbf{2}}\textbf{O}_\textbf{2}\textbf{(g)}\xrightarrow{}\text{\textbf{N}}_\textbf{2}\textbf{(g)}+\textbf{CO}_\textbf{2(g)}+\textbf{H}_\textbf{2}\textbf{O(l)}$$

Ans. Enthalpy change for a reaction (ΔH) is given by the expression,

ΔH = ΔU + ΔngRT

Where,

ΔU = Change in internal energy

Δng = Change in number of moles of the gaseous
substances in the reaction

For the given reaction,

Δng= Δng (products) – Δng (reactants)

= (2 – 2.5) moles

Δng = – 0.5 moles

Given that,

ΔU = –742.7 kJ mol–1

T = 298 K

R = 8.314 × 10–3 kJ mol–1 K–1

Substituting these values in the expression of ΔH:

ΔH = ΔU + ΔngRT

ΔH = (–742.7 kJ mol–1) + (– 0.5 mol) (298 K) (8.314 × 10–3 kJ mol–1 K–1)

ΔH = –743.9 kJ mol–1

6.9. Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol–1 K–1.

Ans. Heat required to raise the temperature can be calculated from the formula.

q = n × C × ΔT

Given: Mass of Al = 60.0 g

$$\text{n}=\frac{\text{Mass}}{\text{Molar mass}}\\\=\frac{60.0}{27\space\text{g mol}^{-1}}$$

C = 24 J mol–1 K–1

ΔT = Tf – Ti

= (55 + 273 – 35 – 273) K

= (328 – 308) K = 20 K

Putting the values in the above relation,

$$q=\bigg(\frac{60}{27}\space\text{mol}\bigg)×(24 \space\text{J} \text{mol}^{\normalsize-1}K^{\normalsize-1})(20 \text{K})$$

= 1066.7 J = 1.07 kJ

Thus, the heat required in kJ is 1.07 kJ.

6.10. Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C.

ΔfusH = 6.03 kJ mol–1 at 0°C.

Cp [H2O(l)] = 75.3 J mol–1 K–1

Cp [H2O(s)] = 36.8 J mol–1 K–1

Ans. Total enthalpy change involved in the transformation is the sum of the following changes and can be represented as:

According to Hess’s law:

(ΔH) = (ΔH1) + (ΔH2) + (ΔH3)

(ΔH1) = (75.3 J mol–1 K–1) (0 – 10)K

= –753 J mol–1

(ΔH2) = ΔHfusion = –6.03 × 103 J mol–1)

(ΔH3) = (36.8 J mol–1 K–1) (–10 – 0)K

= –368 J mol–1

Substituting the values we get,

= –753 J mol–1 – 6030 J mol–1 – 368 J mol–1

= –7151 J mol–1

= –7.15 kJ mol–1

Hence, the enthalpy change involved in the transformation is –7.15 kJ mol–1.

6.11. Enthalpy of combustion of carbon to CO2v is –393.5 kJ mol–1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas.

Ans. Formation of CO2 from carbon and dioxygen gas is represented as:

$$\underset{(1 mole = 44 g)}{\text{C(s)+2O}_2\text{(g)}}\xrightarrow{}2\text{CO}_2\text{(g)}\space\Delta f_c\text{H}=-393.5\text{\space kJ\space\text{mol}}^{\normalsize-1}$$

Heat released on formation of 44 g CO2= –393.5 kJ mol–1

Therefore, heat released on formation of 35.2 g

$$\text{CO}_2=-\frac{393.5×35.2}{44}=-314.8\space\text{kJ\space\text{mol}}^{\normalsize-1}$$

6.12. Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4 (g) are –110 kJ mol–1, –393 kJ mol–1 , 81 kJ mol–1 and 9.7 kJ mol–1 respectively. Find the value of ΔrH for the reaction:
N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)

Ans. ΔrH for a reaction is calculated as the difference between ΔfH (Heat of formation) value of products and ΔfH (Heat of formation) value of reactants.

For the given reaction,

N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)

Substituting the values of ΔfH for N2O, CO2, N2O4, and CO from the question, we get:

ΔrH = [{ ΔfH (N2O) + 3 ΔfH(CO2)}] – [{ΔfH (N2O4)
+ 3 ΔfH (CO) }]

= [{(81) + 3 (– 393)}] – [{(9.7) + 3 (– 110 }] kJ mol–1

= – 777.7 kJ mol–1

6.13. Given

N2(g) + 3H2(g) → 2NH3(g) ΔrH = –92.4 kJ mol–1

What is the standard enthalpy of formation of NH3 gas?

Ans. Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state.

For 1 mole of NH3(g),

$$\frac{1}{2}\text{N}_2\text{(g)}+\frac{3}{2}\text{H}_2\text{(g)}\xrightarrow{}\text{NH}_3\text{(g)}$$

Standard enthalpy of formation of NH3(g)

$$\frac{1}{2}∅_r\text{H}\\=\frac{1}{2}\space(-92.4\space\text{kJ}\space\text{mol}^{\normalsize -1})\\= –46.2 \text{kJ mol}^{–1}$$

6.14. Calculate the standard enthalpy of formation of CH3OH(l) from the following data:

$$\textbf{CH}_\textbf{3}\textbf{OH}\textbf{(l)}+\frac{\text{3}}{\text{2}}\textbf{O}_\textbf{2}\textbf{(g)}\xrightarrow{}\textbf{CO}_\textbf{2}\textbf{(g)}+\textbf{2H}_\textbf{2}\textbf{O(l)} ;\Delta_\text{r}\text{H}^⊖=-726\textbf{kJ mol}^{\normalsize-1}\\\textbf{C}\textbf{(graphite)}+\textbf{O}_2\textbf{(g)}\xrightarrow{}\textbf{CO}_2\textbf{(g)}; \space\Delta_\textbf{c}\textbf{H}^⊖=-393\textbf{kJ mol}^{\normalsize-1}\\\textbf{H}_\textbf{2}\textbf{(g)} +\frac{\textbf{1}}{\textbf{2}}\textbf{O}_\textbf{2}\textbf{(g)}\xrightarrow{}\text{H}_2\textbf{O}(l);Δ_f\text{H}^⊖=-286\textbf{kJ mol}^{\normalsize-1}$$

Ans. Given,

$$\text{CH}_\text{3}\text{OH}\text{(l)}+\frac{\text{3}}{\text{2}}\text{O}_\text{2}\text{(g)}\xrightarrow{}\text{CO}_\textbf{2}\text{(g)}+\text{2H}_\text{2}\text{O(l)} ;\Delta_\text{r}\text{H}^-=-726\text{kJ mol}^{\normalsize-1}\text{...(i)}\\\text{C}\text{(graphite)}+\text{O}_2\text{(g)}\xrightarrow{}\text{CO}_2\text{(g)}; \space\Delta_\text{c}\text{H}^-=-393\text{kJ mol}^{\normalsize-1}\text{...(ii)}\\\text{H}_\text{2}\text{(g)} +\frac{\text{1}}{\text{2}}\text{O}_\text{2}\textbf{(g)}\xrightarrow{}\text{H}_2\text{O}(l);Δ_f\text{H}^-=-286\text{kJ mol}^{\normalsize-1}\text{...(iii)}$$

The reaction that takes place during the formation of CH3OH(l) can be written as:

$$\text{C(s)}+2\text{H}_2\text{O (g)}+\frac{1}{2}\text{O}_2\text{(g)}\xrightarrow{}\text{CH}_3\text{OH}(l)\space...(1)$$

The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:

Equation (ii) + 2 × equation (iii) – equation (i)
fH[CH3OH(l)] = ∆cH + 2∆fH [H2O(l)] – ∆rH
= (– 393 kJ mol–1) + 2(– 286 kJ
mol–1) – (– 726 kJ mol–1)
= (– 393 – 572 + 726) kJ mol–1
∴ ∆fH[CH3OH(l)] = – 239 kJ mol–1

6.15. Calculate the enthalpy change for the process

$$\text{CCl}_4\text{(g)} \xrightarrow{} \text{C(g)} + 4\text{Cl(g)}$$

and calculate bond enthalpy of C – Cl in CCl4(g).

ΔvapH (CCl4) = 30.5 kJ mol–1

ΔfH(CCl4) = –135.5 kJ mol–1

ΔaH(C) = 715.0 kJ mol–1, where ΔaH is enthalpy of atomisation

ΔaH(Cl2) = 242 kJ mol–1

Ans. The chemical equations implying to the given values of enthalpies are:

$$\text{(i)}\space\text{CCl}_4(l)\xrightarrow{}\space\text{CCl}_4\text{(g)}\space\Delta_{\text{vap}}\text{H}^{⊖}=30.5 \text{kJ mol}^{\normalsize–1}\\\text{(ii) C(s)}\xrightarrow{}\text{C(g)}\space\Delta_a\text{H}^{⊖}=715.0\text{kJ mol}^{\normalsize-1}\\\text{(iii) Cl}_2\text{(g)}\xrightarrow{}\Delta_a\text{H}^{⊖}=242\space\text{kJ mol}^{\normalsize-1}\\\text{(iv) C(g) + 4Cl(g)}\xrightarrow{}\text{CCl}_4\text{(g)}\Delta_f\text{H}=-135.5\text{kJ\space\text{mol}}^{\normalsize-1}$$

The required equation is

$$\text{CCl}_4\text{(g)}\xrightarrow{}\text{C(g)}+4\text{Cl(g)}$$

and it can be obtained by multiplying equation (iii) by 2 and adding it to equation (ii) and then subtracting equation (i) and (iv). It can be given as :

Equation (ii) + 2 × equation (iii) – equation (i)– equation (iv)

ΔH = ΔaH(C) + 2(ΔaH(Cl2)) – ΔvapH – ΔfH

= (715.0 kJ mol–1) + 2(242 kJ mol–1)– (30.5 kJ mol–1) – (–135.5 kJ mol–1)

∴ ΔH = 1304 kJ mol–1

Bond enthalpy of C—Cl bond in CCl4(g)

$$=\frac{1304}{4}\text{kJ\space\text{mol}}^{\normalsize-1}$$

= 326 kJ mol–1

6.16. For an isolated system, ΔU = 0, what will be ΔS?

Ans. As change in internal energy ΔU for an isolated system is 0, it means that the exchange of energy with the surroundings is not taking place. In such process, ΔS will be positive and the reaction will be spontaneous. This means ΔS is greater than zero.

Therefore, ΔS > 0 or positive.

6.17. For the reaction at 298 K,

$$\text{2A + B}\xrightarrow{}\text{C}$$

ΔH = 400 kJ mol–1 and ΔS = 0.2 kJ K–1 mol–1

At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?

Ans. We know that,

ΔG = ΔH – TDS

Assuming the reaction at equilibrium, ΔT for the reaction would be:

$$\text{T}=\Delta\text{H}-\frac{\Delta \text{G}}{\Delta \text{S}}\\\text{T}=\frac{∅\text{H}}{∅\text{S}}\space\text{(when,} \Delta \text{G = 0 at equilibrium)}\\\text{T}=\frac{400}{0.2}$$

T = 2000 K.

Thus, the reaction will be in a state of equilibrium at 2000 K and spontaneous above this temperature.

6.18. For the reaction, 2Cl(g) → Cl2(g), what are the signs of ΔH and ΔS ?

Ans. According to the question, 1 mole of chlorine molecule is formed from chlorine atoms. Here, bond formation is taking place, therefore, the energy is released. Hence, ΔH is negative.

If entropy is considered, one mole of molecule atoms have less randomness than two moles of atoms. In the above reaction, the atoms are converted into molecule so, entropy is decreased, hence ΔS is negative for the given reaction. Thus, for the given reaction, the signs of both ΔH and ΔS are negative.

6.19. For the reaction

$$\textbf{2A(g) + B(g)}\xrightarrow{}\textbf{2D(g)}$$

ΔU = –10.5 kJ and ΔS = –44.1 JK–1.

Calculate ΔG for the reaction, and predict whether the reaction may occur spontaneously.

Ans. For the given reaction,

2 A(g) + B(g) → 2D(g)

Δng = 2 – (3) = –1 mole

ΔU = – 10.5 kJ

R = 8.314 × 10–3 kJ–1mol–1

T = 298 K

Substituting the value of ΔU ng, R and T in the expression of ΔH:

ΔH = ΔU + ΔngRT

= (–10.5 kJ) + [(–1 mol) × (8.314 × 10–3 kJ
K–1 mol–1) (298 K)

= –10.5 kJ – 2.478 kJ

= –12.978 kJ

When we substitute the values of ΔH and ΔS in the expression of ΔG, then according to Gibbs’ Helmholtz equation;

ΔG = ΔH – TDS

= (–12.978 kJ) – (298 K) × (–0.0441 kJ
K–1)

= –12.978 + 13.112 = – 12.978 + 13.142

= 0.164 kJ

Since ΔG for the reaction is positive, the reaction will not occur spontaneously.

6.20. The equilibrium constant for a reaction is 10. What will be the value of ΔG? R = 8.314 JK–1mol–1, T = 300 K.

Ans. Given : Using the expression:

ΔG = –2.303 RT log K

ΔG = –(2.303) (8.314 JK–1 mol–1) (300 K) log10

= –5744.14 Jmol–1

= –5.744 kJ mol–1

6.21. Comment on the thermodynamic stability of NO(g), given

$$\frac{1}{2}\textbf{N}_2\textbf{(g)}+\frac{1}{2}\textbf{O}_2\textbf{(g)}\xrightarrow{}\textbf{NO(g)}; \Delta_\textbf{r}\textbf{H}^⊖=\textbf{90\space kJ mol}^{\normalsize-1}\\\textbf{NO(g)}+\frac{1}{2}\textbf{O}_2\textbf{(g)}\xrightarrow{}\textbf{(NO)}_2\textbf{(g);}\Delta_\textbf{r}\textbf{H}^{⊖}=\textbf{–74 kJ mol}^{\normalsize–1}$$

Ans. The positive value of ΔrH indicates the absorption of heat during the formation of NO(g) which means that NO(g) has higher energy than the reactants (N2 and O2). Therefore, NO(g) is unstable.

Moreover, the negative value of ΔrH indicates that heat is evolved during the formation of NO2(g) from NO(g) and O2(g). The minimum energy of the product, NO2(g) makes it a stable one. Thus, unstable NO(g) changes to stable NO2(g).

Hence,

For NO(g); ΔrH = +ve : Unstable in nature

For NO2(g); ΔrH = –ve : Stable in nature

6.22. Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ΔfH = –286 kJ m.

Ans. Given that,

ΔfH = –286 kJ m.

This means that 286 kJ mol–1 of heat is evolved in the formation of 1 mol of H2O(l). Thus, an equal amount of heat will be absorbed by the surroundings.

qrev = (–ΔfH) = –286 kJ mol–1

= 286000 J mol–1

$$\Delta_s\text{(surroundings)}=\frac{\text{q}_{rev}}{\text{T}}=\frac{(286000 \text{J mol)}^{\normalsize-1}}{298 \text{K}}$$

= 959 JK–1 mol–1